Verified answer

Hi!

Since we know the the acceleration we know the rate at which velocity changes should be in equal time intervals

V= 97 km/h or 26.94 m/s

V0= 82 km/h or 22.78 m/s

a = 0.81 m/s^2

now using the formula

v^2 -v0^2 / 2a = d or x

(26.94)^2 - (22.78)^2 / 2(0.81) = x

x = 127.68 meters approximately

therefore distance travelled is 1.3*10^2 meters if it is in 2 sig figs.

Btw just incase you didn't understand how i got that formula

What i would do is first find the t

t = v-v0/a

and substitute that to find distance. We know that distance is (v1+v2)/2 * t

so instead of t i would substitute this

(v+v0)(v-v0/a)/2

which would be v^2-v0^2/2a = d or x

Hope that helps! Its just deriving the formula

And incase your not comfortable with the conventions for the variables i have used

v = velocity final

v0 = initial velocity

d or x = displacement

a = acceleration!

hope this helps! Good luck.

eighty one Km/h = eighty one * a million,000 / 3600 = 22.5 m/s ninety 4 Km/h = ninety 4 * a million,000 / 3600 = 26.11 m/s Vt^2 = V0^2 + 2 * a * S 26.11^2 = 22.5^2 + 2 * 0.86 * S 681.seventy 3 = 506.25 + a million,72 * S a hundred seventy five.40 8 = a million.72 * S S = 102 m

82 kph = 22.7777778 m/s

97 kph = 26.9444444 m/s

Time taken = (26.9444444 - 22.7777778) / 0.81

= 5.14s

Distance = 22.78*5.14 + 0.5*0.18*5.14^2 = 119.47